Where Does e^(2*pi*i) Originate?

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The discussion centers on the origin of the term e^(2*pi*i) in a textbook, with participants explaining that it relates to the periodic nature of the exponential function in complex analysis. They clarify that e^(2*pi*i) equals 1 due to Euler's formula, where e^(ix) = cos(x) + i sin(x), and that adding 2*pi*i does not change the value of e^z. The term is considered a useful identity, reflecting the periodicity of the exponential function in the complex plane. Participants express frustration over the lack of explanation in the textbook, emphasizing the importance of understanding the context of such mathematical properties. Ultimately, the term serves to illustrate a significant concept in Euler's work.
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Homework Statement


The problem at hand is that I don't understand wherefrom my textbook got a certain term(e^(2*pi*i). It doesn't say. At least not as I understand it.
The book says:

Homework Equations


e^(z+2*pi*i) = e^z*e^(2*pi*i) = e^z*1 = e^z

From where does e^(2*pi*i) come? I get the stuff leading to the answer, I just can't seem to understand from where that term comes from.
 
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Basic property of exponents:

e^{a+b} = e^a e^b
 
I'm not certain as to what exactly your asking, but I hope this helps!

Recall the identity property of exponents:

ea+b=eaeb

In your particular case, let:
a=z
b=2*pi*i

Therefore,
ez+(2*pi*i) = ez*e2*pi*i
 
Oh, no, I am sorry if I was not clear. I simply don't know wherefrom they get the 2*pi*i from in e^(z+2*pi*i).

The information I get is what I've written. I believe that the 2*pi refers to the period. It just seems kind of abrupt to randomly insert it without any proof or reference to hardly anything..
 
From calculus one learns that e^{i\pi} = -1
So, using a certain property of exponents, e^{2i\pi} = (e^{i\pi})^2 = (-1)^2 = 1
 
OK. Could be they just added 2*pi*i at random. Why? Because they can :wink:

Do you know Euler's formula?

e^{ix} = \cos x + i \sin x

If you combine it all you see if you insert 2*pi*i into exponent at random, you will not change the result. Sometimes it can be a useful identity.
 
Actually they could add any integer multiple of 2\pi i and still leave the answer unchanged.

e^z=e^{z+2\pi i n} where n is any integer.
 
sirwalle said:
I just can't seem to understand from where that term comes from.
We can't directly help you if you don't show us what they were doing up to that point.
 
Mentallic said:
Actually they could add any integer multiple of 2\pi i and still leave the answer unchanged.

Lol, it must have been a senior moment on my side. I intended to write 2*pi*i*n but looked at 2*pi*i and decided there already is an integer (i) in the formula
 
  • #10
Borek said:
Lol, it must have been a senior moment on my side. I intended to write 2*pi*i*n but looked at 2*pi*i and decided there already is an integer (i) in the formula

People seem to find new uses for i each and every day :wink:
 
  • #11
Hurkyl said:
We can't directly help you if you don't show us what they were doing up to that point.

That's the thing. They aren't doing anything, it has its own little "information box". It says nothing after, nothing before. Just what I've written. All I know is that it has to do with Euler (the chapter is about Euler), if that helps?
 
  • #12
So it must be what I told you earlier - they just show an interesting and important property.

It is like asking where did the 2*pi came from in sin(x+n*2*pi) = sin(x) :-p
 
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